601. Решите уравнение: a) \frac{2x - 5}{x + 5} - 4 = 0; б) \frac{12}{7 - x} = x; в) \frac{x^2 - 4}{4x} = \frac{3x - 2}{2x}; г) \frac{10}{2x - 3} = x - 1; д) \frac{8}{x} = 3x + 2; е) \frac{x^2 + 4x}{x+2} = \frac{2x}{3}; ж) \frac{2x^2 - 5x + 3}{10x - 5} = 0; з) \frac{4x^3 - 9x}{x + 1,5} = 0.

a) \(\frac{2x - 5}{x + 5} - 4 = 0\)

ОДЗ: \(x
eq -5\)

\(\frac{2x - 5 - 4(x+5)}{x + 5} = 0\)

\(\frac{2x - 5 - 4x - 20}{x + 5} = 0\)

\(\frac{-2x - 25}{x + 5} = 0\)

\(-2x - 25 = 0\)

\(-2x = 25\)

\(x = -12.5\)

б) \(\frac{12}{7 - x} = x\)

ОДЗ: \(x
eq 7\)

\(12 = x(7 - x)\)

\(12 = 7x - x^2\)

\(x^2 - 7x + 12 = 0\)

По теореме Виета:

\(x_1 + x_2 = 7\)

\(x_1 \cdot x_2 = 12\)

\(x_1 = 3\)

\(x_2 = 4\)

в) \(\frac{x^2 - 4}{4x} = \frac{3x - 2}{2x}\)

ОДЗ: \(x
eq 0\)

\(\frac{x^2 - 4}{2} = 3x - 2\)

\(x^2 - 4 = 6x - 4\)

\(x^2 - 6x = 0\)

\(x(x - 6) = 0\)

\(x_1 = 0\) не подходит, т.к. не входит в ОДЗ

\(x_2 = 6\)

г) \(\frac{10}{2x - 3} = x - 1\)

ОДЗ: \(x
eq \frac{3}{2}\)

\(10 = (x - 1)(2x - 3)\)

\(10 = 2x^2 - 3x - 2x + 3\)

\(2x^2 - 5x - 7 = 0\)

\(D = (-5)^2 - 4 \cdot 2 \cdot (-7) = 25 + 56 = 81\)

\(x_1 = \frac{5 + \sqrt{81}}{2 \cdot 2} = \frac{14}{4} = \frac{7}{2} = 3.5\)

\(x_2 = \frac{5 - \sqrt{81}}{2 \cdot 2} = \frac{-4}{4} = -1\)

д) \(\frac{8}{x} = 3x + 2\)

ОДЗ: \(x
eq 0\)

\(8 = x(3x + 2)\)

\(8 = 3x^2 + 2x\)

\(3x^2 + 2x - 8 = 0\)

\(D = 2^2 - 4 \cdot 3 \cdot (-8) = 4 + 96 = 100\)

\(x_1 = \frac{-2 + \sqrt{100}}{2 \cdot 3} = \frac{8}{6} = \frac{4}{3}\)

\(x_2 = \frac{-2 - \sqrt{100}}{2 \cdot 3} = \frac{-12}{6} = -2\)

е) \(\frac{x^2 + 4x}{x+2} = \frac{2x}{3}\)

ОДЗ: \(x
eq -2\)

\(3(x^2 + 4x) = 2x(x+2)\)

\(3x^2 + 12x = 2x^2 + 4x\)

\(x^2 + 8x = 0\)

\(x(x + 8) = 0\)

\(x_1 = 0\)

\(x_2 = -8\)

ж) \(\frac{2x^2 - 5x + 3}{10x - 5} = 0\)

ОДЗ: \(x
eq \frac{1}{2}\)

\(2x^2 - 5x + 3 = 0\)

\(D = (-5)^2 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1\)

\(x_1 = \frac{5 + \sqrt{1}}{2 \cdot 2} = \frac{6}{4} = \frac{3}{2}\)

\(x_2 = \frac{5 - \sqrt{1}}{2 \cdot 2} = \frac{4}{4} = 1\)

\(x_1 = \frac{3}{2}\) не подходит, т.к. не входит в ОДЗ

з) \(\frac{4x^3 - 9x}{x + 1,5} = 0\)

ОДЗ: \(x
eq -1.5\)

\(4x^3 - 9x = 0\)

\(x(4x^2 - 9) = 0\)

\(x_1 = 0\)

\(4x^2 - 9 = 0\)

\(4x^2 = 9\)

\(x^2 = \frac{9}{4}\)

\(x_2 = \frac{3}{2}\)

\(x_3 = -\frac{3}{2}\) не подходит, т.к. не входит в ОДЗ

Ответ: a) \(x=-12.5\); б) \(x_1=3, x_2=4\); в) \(x=6\); г) \(x_1=3.5, x_2=-1\); д) \(x_1=\frac{4}{3}, x_2=-2\); е) \(x_1=0, x_2=-8\); ж) \(x=1\); з) \(x_1=0, x_2=\frac{3}{2}\)